Immunology Chapter 5

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  1. Define somatic recombination vs. meiosis and somatic mutation vs. germline mutation
    • 1. somatic recombination - DNA recombination that takes place in somatic cells and not germ cells
    • 2. somatic mutation - DNA mutation that takes place in somatic cells and not germ cells
  2. Describe the construction of light chain genes from gene segments.
    • 1. A variable (V) and a joining (J) gene segment in the genomic DNA are joined to form a complete light-chain V-region exon via somatic recombination.
    • 2. The V gene segment is preceded by an exon encoding a leader peptide (L)
    • 3. The light-chain C regions is encoded in a separate exon
    • 4. Transcription followed by intron splicing results in mRNA consisting of LVJC regions
    • 5. Translation produces VL and CL
  3. Describe the construction of heavy chain genes from gene segments.
    • 1. Diversity (D) and joining (J) gene segments join via somatic recombination
    • 2. Variable (V) joins DJ via somatic recombination forming a complete VH exon
    • 3. The heavy-chain C region gene is encoded by several exons
    • 4. Transcription followed by intron splicing results in mRNA consisting of LVDJC regions
    • 5. Translation
  4. How does the assembly of Ig light and heavy chains result in the repertoire of V regions?
    There are multiple copies of each gene segment in the genome
  5. What are the relative positions of the V, J, D, and C gene segments in relation to each other for the two light chain loci and the heavy chain locus.
    • 1. The lambda light-chain locus (chromosome 22) has between 29 and 32 V segments and 4-5 pairs of J-C segments
    • 2. The kappa light-chain locus (chromosome 2) has ~30 V segments, a cluster of 5 J segments and one C segments
    • 3. The heavy-chain locus (chromosome 14) has ~40 L/V pair segments, a cluster of 23 D segments, a cluster of 6 J segments and one C segment
  6. How do the recombination signal sequences (RSS) function (the 12/23 rule) to assemble the gene segments of the Ig chains?
    • 1. DNA rearrangement is guided by conserved noncoding DNA sequences (RSS - recombination signal sequences) adjacent to the recombination points flanking the V, D, and J segments.
    • 2. RSS's consist of a heptamer (contiguous with the coding sequence) followed by a spacer (12-23 bps) followed by a nonamer (heptamer-spacer-nonamer motif)
    • 3. V gene segments may be in either forward or reverse transcriptional orientation relative to downstream gene segments
    • 4. When a forward-oriented V gene segments recombines with a downstream gene segment, alignments on the two RSS regions loops out the intervening DNA
    • 5. After recombination this loop is excised from the chromosome, taking the two RSS regions with it
    • 6. When a reverse-oriented V gene segment recombines with a downstream gene segment, alignment of the RSS regions forms the intervening DNA into a coiled configuration
    • 7. After recombination the coiled region is retained in the chromosome in an inverted orientation
  7. What is the basic role of the RAG1/2 enzymes in producing coding joints and signal joints (The molecular mechanism of V-region rearrangement).
    • 1. RAG 1/2 (recombination-activation genes) complex binds RSS
    • 2. The complex recruits the other RSS and forms the DNA loop
    • 3. RAG endonuclease activity makes ss cut in DNA right at between the coding segement and the RSS. The 3'OH on the cut strand binds with the 5'phosphodiester bond on the other strand to form a hairpin (coding joint) and leaves a blunt ds break on the RSS (signal joint)
    • 4. Ku70:Ku80 complex (repair protein) binds the DNA on both the coding joints and the signal joints
    • 5. Signal joint - DNA ligase IV:XRCC4 complex ligates the blunt DNA ends forming a precise signal joint
    • 6. Coding joint - DNA-PK:Artemis (protein kinase catalyst:nuclease) complex randomly opens the DNA hairpins leaving either two strands or a single strand extension
    • 7. TdT (Terminal deoxynucleotidyl transferase - adds nucleotides) processes the DNA ends
    • 8. DNA ligase IV:XRCC4 complex ligates the DNA ends forming an imprecise coding joint
  8. Define combinatorial diversity and junctional diversity.
    • 1. combinatorial diversity - V region diversity due to the multiple copies of V genes combining in multiple arrangements
    • 2. junctional diversity - V region diversity due to the diversity at the joints of the segments resulting from the addition and subtraction of nucleotides in the recombination process
  9. Differentiate between P nucleotides and N nucleotides.
    • 1. RSSs are brought together by RAG 1/2 complex
    • 2. RAG complex generates DNA hairpins at coding ends
    • 3. DNA-PK:Artemis complex creates a palindromic sequence (a stretch of nucleotides originating from the complementary strand - P-nucleotides) by converting the hairpin to a ss tail by cleaving the DNA at a point different from the initial break induced by RAG 1/2
    • 4. TdT randomly adds N-nucleotides to the ss strands
    • 5. The two ss base pair
    • 6. Exonuclease - removes unpaired nucletodes
    • 7. DNA synthesis and ligation fill in the gaps to form the coding joint.
  10. Why do CDR1 and CDR2 have less hypervariability than CD3
    • 1. CDR1, 2 are coded w/in the V gene segment
    • 2. CDR3 falls at the joint between the V gene segment and the J gene segment
    • 3. The diversity is increased by the addition of P- and N-nucleotides and the subtraction of nucleotides
  11. Describe the germline organization of the human TCR alpha and beta loci.
    • 1. The variable (V), diversity (D), joining(J), and constant (C) gene segments are distanced apart
    • 2. The TCR alpha locus (chromosome 14) consists of a leader sequence, a cluster of 70-80 V segments, a cluster of 61 J segments, and a C gene
    • 3. The TCR beta locus (chromosome 7) consists of a leader sequence, a cluster of 52 V segments, a repeated cluster of D, 6 J segments, and a C gene.
  12. Describe the germline organization of the TCR gamma:delta chain loci.
    • 1. The alpha chain and delta chain loci consists of a leader sequence, a cluster of 70-80 V alpha and V delta regions interspersed, a cluster of 3 D delta segments, a cluster of 4 J delta segments, a C delta gene, a cluster of 61 J alpha segments, and a C alpha gene
    • 2. The gamma chain locus consists of a leader sequence, a cluster of 12 V segments, a cluster of 3 J segments a C gene, a cluster of 2 J segments, and another C gene
  13. What is the relative abundance of the Ig classes?
  14. What is the MW of Ig classes?
    (146, 146, 165, 146,) 970, (160, 160,) 184, 188
  15. What are the heavy chains that define the Ig classes
    (gamma 1-4), Mu, (alpha 1, 2), delta, epsilon
  16. What is the capability of complement activation for the Ig classes?
    (++, +, +++, -), ++++, (-, -,) -, -
  17. What is the capability of placental transfer for the Ig classes?
    (+++, +, ++, +/-) -, (-, -), -, -
  18. Why are mature B cells able to co-express IgM and IgD?
    • 1. Transcription beginning at the VH promoter extends through C-mu and C-delta exons
    • 2. mRNA processing yields either IgM or IgD in the mature cell
  19. How is a B cell able to switch from membrane-bound forms of Ig to secreted forms?
    • 1. Initial mRNA transcript contains both a SC (secretion) region and a MC (membrane) region
    • 2. During post-transcriptional processing
    • 3. If polyadenylation occurs at the 2nd polyA site then the transcript will contain the code for a membrane binding domain and the Ig will be a membrane bound Ig
    • 4. If polyadenylation occurs at the 1st polyA site then the transcript will contain the code for secretion and the Ig will be secreted
  20. What is the role of the J chain in the polymerization of IgA and IgM
    • 1. IgA and IgM are usually constructed as dimers and pentamers respectively
    • 2. The J chain promote polymerization by binding to the cysteine residue tails of the C chains
  21. What are the 1/2 lives (days) of IgG and IgM?
    (21, 20, 7, 21,) 10
  22. What is the concept of affinity maturation?
    • 1. The increase in Ab affinity for its specific Ag over the course of an adaptive response
    • 2. Prominent in secondary and subsequent immunizations
  23. What is the basic mechanism by which somatic hypermutation is accomplished in the Ig V region?
    • 1. AID (activation-induced cytidine deaminase) converts a cytidine (C) to a uridine (U) in an Ig gene
    • 2. The repair mechanism can create point mutations at and around the of the original C:G pair
  24. How does affinity maturation relate to somatic hypermutation?
    • 1. Somatic hypermutation - Extensive mutation that occurs in the V region DNA sequence of rearranged Ig genes in activated B cells
    • 2. Results in the production of variant Igs that bind with greater affinity to their Ags
  25. What is the basic mechanism for class switching
    • 1. Transcription through the switch region (intron) of a heavy chain locus is initiated by activation of the upstream promoter
    • 2. AID, UNG, APE1 introduce clustered nicks on beoth strands of DNA
    • 3. Repair proteins act to initiate double-strand break repair (DSBR)
    • 4. DSBR machinery joins the two switch regions and excises intervening sequences
    • 5. The selected C region is not located adjacent to the VDJ region
  26. What is the rational for the immune system performing class switching
    Class switching enables the same assembled VH exon to be associated with different CH genes in the course of an immune response

Card Set Information

Immunology Chapter 5
2012-02-27 18:29:59
Lymphocyte Ag Receptor Generation

Chapter 5
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